# Doppler shift for signals from an infalling astronaut using Eddington-Finkelstein coordinates

Last year I was given the following question as part of an exam, but looking back I'm not sure how to get the answer.

Q: Alice is an astronaut who falls radially into a Schwarzschild black hole. During her fall, she sends radio signals radially to her friend Bob, who is at constant radius far from the black hole.

(i) Use outgoing Eddington-Finkelstein coordinates $(u, r, \theta, \phi )$ to show that the redshift of the signals received by Bob is given by $$\frac{\lambda_B}{\lambda_A} = \frac{du}{d\tau}$$ where Alice’s trajectory is $u=u(\tau)$, $r=r(\tau)$, $\tau$ is her proper time, and the RHS is evaluated at the time the signal is emitted.

Naive attempt:
$$\frac{\lambda_B}{\lambda_A} = \frac{d\tau_B}{d\tau_A} = \sqrt{1-\frac{2M}{r_B}} \frac{du}{d\tau_A} \quad ???$$

Assuming $\frac{du}{d\tau_A}$ is equal to the RHS desired, I have an extra square root factor unfortunately. Now that I think of it though, surely the solution must have some dependence on $r_B$... do you think this was a typo in the exam?

The outgoing E-F metric is (units of $G=1$, $c=1$, radial motion only) $$d\tau^2 = \left(1 - \frac{2M}{r}\right)\ du^2 + 2du\ dr$$
For Bob, $dr=0$, so $$\Delta \tau_B = \left(1 - \frac{2M}{r_{B}} \right)^{1/2} \Delta u$$ and I think you are supposed to assume that "far from the black hole" means $r_B \gg 2M$, so $\Delta \tau_B = \Delta u$.
Now here, $\Delta \tau_B$ is the proper time between pulses or wavepeaks in Bob's frame and $\Delta u$ is the corresponding change in the $u$ coordinate. But an advantage to the Eddington-Finkelstein coordinates is that outgoing light rays travel on geodesics of constant $u$, and so the value of $\Delta u$ (but not $\Delta \tau$) is exactly the same in Alice's frame.
Thus in Alice's frame $$\Delta \tau_A = \left(\frac{du}{d\tau} \right)^{-1} \Delta u$$
To work out the redshift we use the formula as per your question $$\frac{\lambda_B}{\lambda_A} = \frac{\Delta \tau_B}{\Delta \tau_A} = \frac{\Delta u}{\Delta u (du/d\tau)^{-1}} = \frac{du}{d\tau}$$